all nodes distance k in binary tree

Art Scpae

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19-10-2024

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Finding All Nodes at a Distance K in a Binary Tree: A Comprehensive Guide

Finding nodes in a binary tree at a specific distance from a given node is a common problem in data structures and algorithms. This article will delve into the efficient solution for this problem, exploring the logic, implementation, and practical applications.

Understanding the Problem

Given a binary tree and a target node, we need to find all nodes that are exactly k distance away from the target node. This means considering nodes both upwards (towards the root) and downwards (towards the leaves) in the tree.

The Algorithm

The most efficient approach to solve this problem utilizes a depth-first search (DFS) algorithm combined with a level-order traversal. Here's a breakdown of the steps:

1. Target Node Identification: Start by performing a DFS traversal to locate the target node within the tree.
2. Distance Calculation: From the target node, initiate a level-order traversal to find nodes at the desired distance k.
3. Handling Ancestors: To account for nodes k distance upwards from the target, during the initial DFS traversal, store the distance from the root node for each node. This allows us to calculate the relative distance between the target node and its ancestors.

Code Example (Python):

from collections import defaultdict, deque

class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

def find_nodes_at_distance_k(root, target, k):
    """
    Finds all nodes at a distance K from the target node in a binary tree.

    Args:
        root: The root node of the binary tree.
        target: The target node.
        k: The desired distance.

    Returns:
        A list of nodes at a distance K from the target node.
    """

    # Dictionary to store distance from root for each node.
    distance_from_root = defaultdict(int)
    # Find target node and its distance from root.
    find_target_node(root, target, 0, distance_from_root)
    
    # Perform a level-order traversal from the target node.
    nodes_at_distance_k = []
    queue = deque([(target, 0)])
    visited = {target}
    while queue:
        node, level = queue.popleft()
        # If we reach the desired distance, add the node to the list.
        if level == k:
            nodes_at_distance_k.append(node)
        # If we need to explore further, add the node's children to the queue.
        if level < k:
            if node.left and node.left not in visited:
                queue.append((node.left, level + 1))
                visited.add(node.left)
            if node.right and node.right not in visited:
                queue.append((node.right, level + 1))
                visited.add(node.right)
    
    # Find ancestors at the correct distance.
    for node in distance_from_root:
        # If the distance between the target and its ancestor is k, add the ancestor.
        if distance_from_root[target] - distance_from_root[node] == k:
            nodes_at_distance_k.append(node)

    return nodes_at_distance_k

def find_target_node(node, target, distance, distance_from_root):
    """
    Performs a DFS traversal to find the target node and update distances from the root.

    Args:
        node: The current node being processed.
        target: The target node.
        distance: The current distance from the root.
        distance_from_root: Dictionary to store distances from the root.
    """
    if not node:
        return False
    
    # Update distance from the root for the current node.
    distance_from_root[node] = distance
    
    # Recursively explore the left and right subtrees.
    if node == target:
        return True
    if find_target_node(node.left, target, distance + 1, distance_from_root) or find_target_node(node.right, target, distance + 1, distance_from_root):
        return True
    return False

# Example usage:
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
target = root.left

k = 2
nodes_at_distance_k = find_nodes_at_distance_k(root, target, k)
print(f"Nodes at distance {k} from {target.val}: {[node.val for node in nodes_at_distance_k]}")

Example Explanation:

In the given example, we are tasked with finding nodes at a distance of 2 from node with value 2 (the target). The output would be [4, 6], as these nodes are exactly two steps away from the target node. The code leverages both level-order and depth-first traversal to achieve this efficiently.

Applications:

This algorithm has various applications in areas like:

• Social Network Analysis: Find users k connections away from a specific user.
• Recommendation Systems: Suggest items k steps away from a user's current interests.
• Network Routing: Identify alternative paths k hops away from the current route.

Conclusion:

Finding all nodes at a distance k in a binary tree is a common problem that can be solved effectively using depth-first and level-order traversal techniques. By leveraging these algorithms, we can efficiently explore the tree structure and identify nodes based on their relative distances from a target node. This technique has diverse applications in various domains, showcasing the versatility and practical value of data structures and algorithms in problem-solving.